Question

The age of a
father 8 yrs
back was 5 times
that
his
his son. After 8 yrs.
will be
8 yrs more than
double
the age of
his son
the age
find their present ages.
​

Answer 1

Answer:

Let the present age of father and his son be X and Y year's.

Before 8 years age of father = (X-8) years

Before 8 years age of son = (Y-8) years

According to question,

(X-8) = 5(Y-8)

X - 8 = 5Y - 40

X - 5Y = -40 + 8

X - 5Y = -32------------(1)

After 8 years age of father = (X+8) years

After 8 years age of son = (Y+8) years

According to question,

(X+8) = 2(Y+8) + 8

X + 8 = 2Y + 16 +8

X + 8 = 2Y + 24

X - 2Y = 24 -8

X - 2Y = 16--------(2)

From equation (1) we get,

X - 5Y = -32

X = -32 + 5Y----------(3)

Putting the value of X in equation (2)

X - 2Y = 16

-32 + 5Y - 2Y = 16

3Y = 16 + 32

3Y = 48

Y = 48/3

Y = 16 years

Putting the value of Y in equation (3)

X = -32+5Y => -32 + 5 × 16 => -32 + 80

X = 48 years

Age of father = X = 48 years

And,

Age of son = Y = 16 years

Hope it helps u!